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Spring compression

  1. Mar 8, 2016 #1
    1. The problem statement, all variables and given/known data
    U2crGsz.png

    2. Relevant equations
    ΔUelastic = 1/2(kxf2) - 1/2(kxi2)
    ΔK + ΔUg + ΔUelastic = 0
    3. The attempt at a solution
    a) I used ΔUelastic = 1/2(kxf2) - 1/2(kxi2) and found xf - 0.126 m.
    b) Should I use ΔK + ΔUg + ΔUelastic = 0? I am unsure where to start for this one. How do I go about setting this up?
     
  2. jcsd
  3. Mar 8, 2016 #2

    PhanthomJay

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    Part a looks good and yes part b setup is good. What is the value of the change in K from initial position of book to final position of book at max spring compression?
     
  4. Mar 8, 2016 #3
    Is it 0 because it is being dropped from rest then stops?
     
  5. Mar 8, 2016 #4

    PhanthomJay

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    Y
    yes. And what is the change in gravitational U from book initial position to its final position?
     
  6. Mar 8, 2016 #5
    It will be 0 - (10 kg)(9.8 m/s^2)(0.25 m), correct?
    I got 0.25 m as my final answer, but should I put it has -0.25 or +0.25 for it to be correct? Or is magnitude fine in this?
     
  7. Mar 8, 2016 #6

    PhanthomJay

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    you cannot ignore the minus sign. But you are not correct here. It is given that the book is 0.25 m above the top of the spring when both are at rest in their initial position. How much is it initially from the spring when the spring will have been compressed its max distance?
     
  8. Mar 8, 2016 #7
    So are you saying that 0 - (10 kg)(9.8 m/s^2)(0.25 m) is not the correct way to calculate the change in gravitational U? How else am I supposed to solve for this when I already have 1 unknown variable?
     
  9. Mar 8, 2016 #8

    PhanthomJay

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    The unknown variable is the max spring compression, x. So if the book is .25 m above the spring before it is compressed, how high is it initially above the spring when the spring is compressed by a distance of x? You can use the unknown variable in that expression.
     
  10. Mar 8, 2016 #9
    But the book is dropped so the spring is compressed... My understanding of your explanation is that the spring compresses when the book is 0.25 above the spring, but in this problem, that is not the case.
     
  11. Mar 8, 2016 #10

    PhanthomJay

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    Remember when you calculated the change in K from initial to final position you used 0 Kinetic Energy in the initial position of the book and then 0 Kinetic Energy in the final position after the spring was compressed? Do the same for initial gravitational U and final gravitational U after the spring is compressed. When calculating gravitational U, you need to establish a reference axis for 0 gravitational U. You should take the reference 0 level for Gravitational U at the level when the spring is compressed its max distance, then calculate the change in gravitational U from start point to end point.
     
  12. Mar 8, 2016 #11
    I think I see what you are saying but I am getting confused I think... I made Ugf the reference axis for for 0 gravitational U when the spring is compressed and the book hit the spring. Ugi from my understanding is when the book is 0.25 m above the spring, so there is potential energy there. Kinetic energy in the final position is 0 since the book hit the spring so it all aligns well I think...
    Can you type some work out to show me what you are talking about? I think I need a visual representation.
     
  13. Mar 8, 2016 #12

    PhanthomJay

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    Take the reference level for Ug as 0 when the spring is compressed. So the initial height of the book when it is dropped is 0.25 + x from that reference level, and the final Ug when the spring is fully compressed is 0. Now calculate the initial Ug from that level and the change in Ug. There will be an x in that expression.
     
  14. Mar 8, 2016 #13
    This is how I set it up:
    Initial Final
    Ki = 0 Kf = 0
    Ugi = mgy = (10 kg)(9.8 m/s^2)(0.25 m) Ugf = 0
    Uelastic = 0 Uelastic = (1/2)k(spring compression)^2
    So x is how much the spring will compress when the book drops, so therefore in Ugi, it would be 0.25 + x instead because you also have to account for that compression distance? But how about Uelastic?
     
  15. Mar 9, 2016 #14

    PhanthomJay

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    yes, so correct your equation above for Ugi
    well initially it is zero and its final Uelastic is just the function of its max x compression distance. Applying the conservation of energy equation, you will now get a quadratic equation to solve for x.
     
  16. Mar 9, 2016 #15
    So everything is correct then? I got 0.25 m as the compression distance.
     
  17. Mar 9, 2016 #16

    PhanthomJay

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    No, that is not at all correct. I don't think you are paying attention. You didn't correct your Ugi equation to denote the height above the reference 0 level ...h is not equal to 0.25 m but rather h = (0.25 +x) m.
     
  18. Mar 9, 2016 #17
    I see what you are saying now. If I set the reference point 0 when it isn't compressed, would it be (0.25-x)?
     
  19. Mar 9, 2016 #18
    Also can I use y for spring compression instead?
     
  20. Mar 9, 2016 #19

    PhanthomJay

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    yes, you can replace x with y for all cases if you wish.
    Would what be that? You can set the reference point 0 for Ug at the uncompressed spring level if you want, but then the initial Ugi would be .25mg and the final Ugf would be a negative value (- mgx), and the change in U from initial to final would be Ugf - Ugi , work that out and don't let the minus signs get the best of you.
     
  21. Mar 9, 2016 #20
    Okay I used your reference point that you mentioned initially and I got 0.40 m as the max distance the spring will compress, is that correct?
     
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